## anonymous one year ago How to exchange the integral order, i can't solve this problem. ∫∫ xy dxdy, D = {(x,y) | -2 <= y <= 4, (y^2-6)/2 <= x <= y+1}

1. phi

Here is a plot of the region, showing a "typical" rectangle as we integrate x first.

2. phi

If we integrate dy first, then we evidently have to break the problem into two regions. from x=-3 to -1, and then from x=-1 to 5

3. phi

from the lower limit (in the original problem) $x= y^2 -6$ we get $y= \pm \sqrt{x+6}$ in the region between x=-3 and x=-1, evidently the lower limit is $$y=- \sqrt{x+6}$$ and the upper limit is $$y= \sqrt{x+6}$$ and we have $\int_{-3}^{-1} \int_{- \sqrt{x+6}}^{+ \sqrt{x+6}} x y\ dy \ dx$ once we get to x=-1 we change the lower limit on y. we originally had x=y+1, from which we get y= x-1 the second iterated integral is $\int_{-1}^{5} \int_{x-1}^{+ \sqrt{x+6}} x y\ dy \ dx$